Extensions of Mendelian Genetics

Question 1 (10 marks)

(a) ABO system involves 3 alleles ($I^A$, $I^B$, $i$) in the population — more than two possible forms = multiple alleles (2). In genotype $I^A I^B$, both A and B antigens are fully expressed simultaneously (neither masks the other) = codominance (2). Why: codominance = both alleles expressed independently; multiple alleles = >2 variants exist though any individual carries only 2.

(b) Man heterozygous type B = $I^B i$ (1). Woman type AB = $I^A I^B$ (1).

(c) Cross $I^A I^B \times I^B i$:

Offspring: 1 AB : 1 A : 2 B → 1 AB : 1 A : 2 B (grid 2, ratio 2).

Question 2 (8 marks)

(a) The $C^R$ allele produces functional enzyme making red pigment; $C^W$ produces none. Heterozygote has only ONE dose of functional allele → half the pigment → intermediate pink phenotype (dosage/haploinsufficiency) (3).

(b) $C^R C^W \times C^R C^W$: genotypes $1\ C^RC^R : 2\ C^RC^W : 1\ C^WC^W$ = 1 red : 2 pink : 1 white (genotype 1.5, phenotype 1.5). Genotypic = phenotypic ratio because heterozygote is distinct.

(c) In incomplete dominance the F₁ shows a blended/intermediate phenotype (pink); in codominance the F₁ shows BOTH parental phenotypes simultaneously (e.g., AB, both antigens) — not a blend (2).

Question 3 (12 marks)

(a) Carrier mother $X^H X^h$ × unaffected father $X^H Y$:

(grid 3). Proportions: affected sons = 1/2 of sons (1/4 total); carrier daughters = 1/2 of daughters (1/4 total); affected daughters = 0 (3).

(b) Males are hemizygous — one X only — so a single recessive allele is expressed (no second X to mask it). Females need two copies (homozygous) to be affected, which is rarer, requiring both an affected/carrier parent contribution (3).

(c) Fathers pass their Y (not X) to sons, so no X-linked allele reaches a son (1.5). But a father passes his $X^h$ to ALL daughters (obligate carriers); a carrier daughter may then pass $X^h$ to her son (grandson) → skipped-generation inheritance (1.5).

Question 4 (10 marks)

(a) Epistasis = one gene at one locus masks/modifies the expression of a gene at a DIFFERENT locus (2). Differs from dominance, which is interaction between ALLELES of the SAME gene/locus (1).

(b) BbEe × BbEe → standard 9:3:3:1 genotype grid:

• 9 B_E_ = black
• 3 bbE_ = brown
• 3 B_ee = yellow
• 1 bbee = yellow

Since ee (any B) = yellow: 3 + 1 = 4 yellow. 9 black : 3 brown : 4 yellow (5).

(c) Modified ratio 9 : 3 : 4; this is recessive epistasis (ee, the recessive homozygote, is epistatic) (2).

Question 5 (12 marks)

(a) RF = (recombinant offspring / total offspring) × 100, measuring crossover frequency, which is proportional to distance between loci (2). By definition 1% RF = 1 cM (map unit) (1).

(b) Order determined by additivity: A–C (17) is largest = A and C are outermost. A–B = 12, B–C = 5, and 12 + 5 = 17 ✓, so B lies between A and C:

A ---12cM--- B ---5cM--- C
   (17 cM total A to C)

(order 2, distances 2).

(c) RF = 52% indicates the genes are either on different chromosomes or so far apart they assort independently — effectively unlinked (2). RF cannot exceed 50% because independent assortment already produces 50% recombinants maximum; multiple crossovers can't raise the observable recombinant fraction above that (1).

(d) Double crossovers between the outer markers go undetected, so the largest measured distance is an underestimate; short-range sums are more accurate (2).

Question 6 (8 marks)

(a) Polygenic = a trait controlled by MANY genes each with additive small effect, producing continuous variation with a bell-shaped (normal) distribution — unlike single-gene discrete Mendelian classes (3).

(b) 3 genes, 2 alleles each → number of additive-dose classes = 2n + 1 = 2(3)+1 = 7 phenotypic classes (2).

(c) Example: hydrangea flower colour changes with soil pH; or temperature affecting Himalayan rabbit fur; or height affected by nutrition (1.5). Norm of reaction = the range of phenotypes a single genotype can produce across different environments (1.5).

[
  {"claim":"Q1c cross IAIB x IBi gives ratio 1 AB : 1 A : 2 B (as counts 1,1,2)","code":"counts={'AB':0,'A':0,'B':0,'O':0}\nman=['IB','i']\nwoman=['IA','IB']\nfor m in man:\n    for w in woman:\n        g=sorted([m,w])\n        if 'IA' in g and 'IB' in g: counts['AB']+=1\n        elif 'IA' in g: counts['A']+=1\n        elif 'IB' in g: counts['B']+=1\nresult = (counts['AB'],counts['A'],counts['B'])==(1,1,2)"},
  {"claim":"Q4b Labrador recessive epistasis gives 9:3:4","code":"from itertools import product\nal={'B':['B','b'],'E':['E','e']}\ncnt={'black':0,'brown':0,'yellow':0}\nfor b1,b2,e1,e2 in product('Bb','Bb','Ee','Ee'):\n    hasB='B' in (b1,b2)\n    hasE='E' in (e1,e2)\n    if not hasE: cnt['yellow']+=1\n    elif hasB: cnt['black']+=1\n    else: cnt['brown']+=1\nresult=(cnt['black'],cnt['brown'],cnt['yellow'])==(9,3,4)"},
  {"claim":"Q5b gene order additive: AB+BC = AC (12+5=17)","code":"result = (12+5)==17"},
  {"claim":"Q6b number of phenotypic classes for 3 additive genes = 7","code":"n=3\nresult = (2*n+1)==7"}
]